Question: Divide the following complex numbers: $\dfrac{6(\cos(\frac{1}{3}\pi) + i \sin(\frac{1}{3}\pi))}{2(\cos(\frac{23}{12}\pi) + i \sin(\frac{23}{12}\pi))}$ (The dividend is plotted in blue and the divisor in plotted in green. Your current answer will be plotted orange.)
Answer: Dividing complex numbers in polar forms can be done by dividing the radii and subtracting the angles. The first number ( $6(\cos(\frac{1}{3}\pi) + i \sin(\frac{1}{3}\pi))$ ) has angle $\frac{1}{3}\pi$ and radius 6. The second number ( $2(\cos(\frac{23}{12}\pi) + i \sin(\frac{23}{12}\pi))$ ) has angle $\frac{23}{12}\pi$ and radius 2. The radius of the result will be $\frac{6}{2}$ , which is 3. The difference of the angles is $\frac{1}{3}\pi - \frac{23}{12}\pi = -\frac{19}{12}\pi$ The angle $-\frac{19}{12}\pi$ is negative. A complex number goes a full circle if its angle is increased by $2 \pi$ , so it goes back to itself. Because of that, angles of complex numbers are convenient to keep between $0$ and $2 \pi$ $-\frac{19}{12}\pi + 2 \pi = \frac{5}{12}\pi$ The radius of the result is $3$ and the angle of the result is $\frac{5}{12}\pi$.